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The commutator of square matrices, $$[A,B]:=AB - BA,$$ can be viewed as a function $[\cdot,\cdot]:\mathbb{R}^{n \times n} \times \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^{n \times n}$ (or $\mathbb{C}^{n \times n} \times \mathbb{C}^{n \times n} \rightarrow \mathbb{C}^{n \times n}$) that takes two matrices as input and returns a matrix as output.

Questions:

  • What is known about the image of this function?
  • What matrices are in the image, and what algebraic and topological structures does the image have?

Stated another way, what are the characteristics of the set of matrices $M$ that can be written in the form $M=AB-BA$ for some $A,B$?

Clearly, a necessary condition is that the trace of a matrix in the image of the commutator must be zero, since $\text{trace}(AB)=\text{trace}(BA)$. However, it is not clear (to me) whether this condition is sufficient to characterize the set (I suspect it is not).

Much information can be found online about the image of the group-theoretic commutator, but I am interested in the ring-theoretic commutator, which is different, and in particular, I am interested in the special case where the commutator is applied to square matrices that are real or complex valued.

Nick Alger
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    Maybe duplicate? – Seewoo Lee Jul 31 '16 at 19:27
  • Notice that anything in the image has trace zero. I'm pretty sure that any traceless matrix is in the image too, but don't have time for a proof at the moment. – Max Jul 31 '16 at 19:55
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    Apparently the answer is that it is precisely the traceless matrices, although the argument's a bit tricky (http://math.stackexchange.com/a/252324/232). A more natural question is about linear combinations of commutators; this always forms a Lie subalgebra, for example, and has other nice properties. – Qiaochu Yuan Aug 01 '16 at 04:32
  • @QiaochuYuan Thanks. Since it turns out that all traceless matrices are commutators (which I did not expect a priori), the answer to that question by user1551 also answers this question. However, what is there to be gained by taking linear combinations of commutators, if they already form a linear subspace by themselves? – Nick Alger Aug 01 '16 at 05:30
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    @Nick: I mean to say that it's always true that if $A$ is a $k$-algebra, then $k$-linear combinations of commutators of $A$ is a Lie $k$-subalgebra of $A$, usually denoted $[A, A]$. It shouldn't be true in general that every element of $[A, A]$ is a commutator, even if it's true in this case; I don't know a counterexample though. The quotient $A/[A, A]$ is called the zeroth Hochschild homology $HH_0(A)$ and is a natural object of study for several reasons. it turns out, for example, to be invariant under Morita equivalence. – Qiaochu Yuan Aug 01 '16 at 06:21

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It turns out that every matrix with zero trace can also be written as a commutator, so the image of the commutator is exactly the set of traceless matrices. See, for example, the excellent answer here by user1551: Does the set of matrix commutators form a subspace?

Also, thanks to Qiaochu Yuan for pointing out this link in the comments.

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Nick Alger
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